2 aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas. how many grams (rounded…

Explanation:

Step1: Write the balanced chemical equation

$2Al + 3H_2SO_4=Al_2(SO_4)_3 + 3H_2$

Step2: Calculate the molar mass of $H_2SO_4$ and $Al_2(SO_4)_3$

The molar mass of $H_2SO_4$: $M_{H_2SO_4}=2\times1 + 32+4\times16=98\ g/mol$. The molar mass of $Al_2(SO_4)3$: $M{Al_2(SO_4)_3}=2\times27 + 3\times(32 + 4\times16)=342\ g/mol$.

Step3: Calculate the number of moles of $H_2SO_4$

$n_{H_2SO_4}=\frac{m_{H_2SO_4}}{M_{H_2SO_4}}=\frac{250\ g}{98\ g/mol}\approx2.551\ mol$.

Step4: Determine the mole - ratio between $H_2SO_4$ and $Al_2(SO_4)_3$

From the balanced equation, the mole - ratio of $H_2SO_4$ to $Al_2(SO_4)3$ is $3:1$. So the number of moles of $Al_2(SO_4)3$ formed, $n{Al_2(SO_4)3}=\frac{1}{3}n{H_2SO_4}$. $n{Al_2(SO_4)_3}=\frac{1}{3}\times2.551\ mol\approx0.8503\ mol$.

Step5: Calculate the mass of $Al_2(SO_4)_3$

$m_{Al_2(SO_4)3}=n{Al_2(SO_4)3}\times M{Al_2(SO_4)_3}=0.8503\ mol\times342\ g/mol\approx290.80\ g$.

Answer:

$290.80\ g$